Chapter 10.1, Problem 10.4E

To determine

To Explain: the chi-square distribution should which used.

Answer to Problem 10.4E

With 5 degrees Chi-square distribution of freedom is $0.01<P<0.05$

Explanation of Solution

Calculation:

The expected frequencies E are the multiplication of the sample size and the probabilities.

  $\begin{array}{l}{E}_1=n{p}_1=46\times 0.23=10.58\\ E_2=n{p}_2=46\times 0.23=10.58\\ E_3=n{p}_3=46\times 0.15=6.9\\ E_4=n{p}_4=46\times 0.15=6.9\\ E_5=n{p}_5=46\times 0.12=5.52\\ E_6=n{p}_6=46\times 0.12=5.52\end{array}$

The chi-square subtotals are the squared differences between the observed and expected frequencies which are divided by the expected frequency.

The value of the test-statistic would be then the sum of the chi-square subtotals:

  $\begin{array}{c}{x}^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}\\ =\frac{{\left(12-10.58\right)}^2}{10.58}+\frac{{\left(7-10.58\right)}^2}{10.58}\\ +\frac{{\left(13-6.9\right)}^2}{6.9}+\frac{{\left(4-6.9\right)}^2}{6.9}\\ +\frac{{\left(8-5.52\right)}^2}{5.52}+\frac{{\left(2-5.52\right)}^2}{5.52}\\ =11.3724\end{array}$

The degrees of freedom is

  $df=c-1=6-1=5$

The P-value is the probability of getting the value of the test statistic, or a value more extreme. The P-value is the number in the column chi-square distribution table containing the $x^2$ -value in the row $df=c-1=6-1=5$ :

  $0.01<P<0.05$